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What is the antiderivative of $(2x+7)^{1/2}$?

My understanding is that it would be $\frac 23 \times(2x+7)^{3/2}$ but according to the source I am working from the answer is $\frac 13 \times (2x+7)^{3/2}$.

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    $\begingroup$ Differentiate your answer and you will see why. HINT: Chain rule. $\endgroup$ – mickep Mar 2 '16 at 10:00
  • $\begingroup$ You probably forgot to process the coefficient of $x$. $\endgroup$ – Yves Daoust Mar 2 '16 at 10:13
  • $\begingroup$ Why do I need to process the coefficient of x before I take the antiderivative? $\endgroup$ – user252704 Mar 2 '16 at 11:57
  • $\begingroup$ Because it is going to be processed when you take the derivative. If it changes in one direction, it has to change in both directions, or else the derivative of an antiderivative wouldn't equal the original function. $\endgroup$ – kevinsa5 Mar 2 '16 at 14:13
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$$\int\sqrt{(2x+7)}dx$$

Set $t=2x+7$ and $dt=2dx$

$$=\frac 1 2\int\sqrt tdt=\frac{t^{3/2}}{3}+\mathcal C=\frac{(2x+7)^{3/2}}{3}+\mathcal C$$

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  • $\begingroup$ Why multiply it by 1/2 and also by 2? $\endgroup$ – user252704 Mar 2 '16 at 12:01
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    $\begingroup$ @user252704 its because $dt=2dx$...therefore $dx=\dfrac{dt}{2}$...now all we need to do is to plug in the values... $\endgroup$ – manshu Mar 2 '16 at 12:39
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Notice:

$$\int x^n\space\text{d}x=\frac{x^{1+n}}{1+n}+\text{C}$$



$$\int\sqrt{2x+7}\space\text{d}x=$$


Substitute $u=2x+7$ and $\text{d}u=2\space\text{d}x$:


$$\frac{1}{2}\int\sqrt{u}\space\text{d}u=\frac{1}{2}\int u^{\frac{1}{2}}\space\text{d}u=\frac{1}{2}\cdot\frac{u^{1+\frac{1}{2}}}{1+\frac{1}{2}}+\text{C}=\frac{u^{\frac{3}{2}}}{3}+\text{C}=\frac{(2x+7)^{\frac{3}{2}}}{3}+\text{C}$$

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  • $\begingroup$ Why multiply it by 1/2 and also by 2? $\endgroup$ – user252704 Mar 2 '16 at 12:01
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    $\begingroup$ To make the substitution clear $\endgroup$ – Jan Mar 2 '16 at 12:59
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$$(2x+7)^{1/2}=2^{1/2}(x+\frac72)^{1/2}\to2^{1/2}\frac1{\frac32}(x+\frac72)^{3/2}=\frac{2^{3/2}}3(x+\frac72)^{3/2}=\frac{(2x+7)^{3/2}}3.$$

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  • $\begingroup$ Why do I need to process the coefficient of x before I take the antiderivative? $\endgroup$ – user252704 Mar 2 '16 at 11:58
  • $\begingroup$ I noticed that you didn't handle it properly so this is a workaround. $\endgroup$ – Yves Daoust Mar 2 '16 at 12:13
  • $\begingroup$ @user252704, it's like the chain rule in reverse... (2x+7)^(1/2) is like f(g), where f(x) = x^(1/2) and g(x) = 2x+7, except here you have to remember to process the ANTIderivative of g instead of the derivative $\endgroup$ – sig_seg_v Mar 2 '16 at 16:50

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